Question

Basically, I’m trying to flatten a list in my function but ignore that (you can also ignore the print functions I put in).
take x = [[1,2,3],4,5] to be my variable.
I call prob7(x) but the issue is that when type([1,2,3]) gets checked == list, it returns false. Why is that? I explicitly check this on the interpreter command line and it returns true. But inside the function, I get a false.

Just a bug that I missed because I’m sleepy or am I misunderstanding some part of the Python language? I run version 2.6 if it matters.

def prob7(list): # flatten a list
    tempList = []
    if list: # meaning if there are elements in the list and it is not empty
        for i in list:
            if type(i) != list:
                print tempList,'if',i,type(i)==list
                tempList.append(i)
            else:
                print tempList,'else',i
                tempList.extend(prob7(i))

    return tempList

Answer 1

Just not use ‘list’ as a variable name and use isinstance(var, list) instead of type(var) == list.
Please find corrected sample below.

def prob7(mylist): # flatten a list
    tempList = []
    if mylist: # meaning if there are elements in the list and it is not empty
        for i in mylist:
            if not isinstance(i, list):
                print tempList, 'if', i, isinstance(i, list)
                tempList.append(i)
            else:
                print tempList, 'else', i
                tempList.extend(prob7(i))    
    return tempList

Or if you don’t really required to use recursion and you don’t care about values order then you can use something like this:

lVals = [[1,2,3],4,5, [1,[4,7]]]

def make_flat(mylist): # flatten a list    
    while any(isinstance(x, list) for x in mylist):
        for i, val in enumerate(mylist):
            if isinstance(val, list):                
                mylist.extend(mylist.pop(i))        
                break
    return mylist

make_flat(lVals)
>>> [4, 5, 1, 2, 3, 1, 4, 7]